Assigning the aldol product from a crude mixture
A student condenses 4-methoxybenzaldehyde (1.36 g, 10.0 mmol) with acetone (2.90 g, 50.0 mmol) in ethanol using aqueous NaOH. After 2 h, the mixture is acidified, extracted with dichloromethane, and concentrated to a yellow oil (1.52 g). TLC in hexane/EtOAc 4:1 shows two major spots: A at Rf 0.62 and B at Rf 0.28. Column chromatography gives A (0.74 g) and B (0.21 g). HRMS of A: m/z 176.0837 [M]+ (calculated for C11H12O2, 176.0837). IR (neat): 1665, 1600, 1510 cm−1. 1H NMR of A (CDCl3): δ 7.50 (d, J = 8.8 Hz, 2H), 7.48 (d, J = 16.2 Hz, 1H), 6.93 (d, J = 8.8 Hz, 2H), 6.60 (d, J = 16.2 Hz, 1H), 3.85 (s, 3H), 2.36 (s, 3H); a small extra doublet at δ 6.55 (d, J = 12.4 Hz) is also visible. B has [M]+ 136.0524 and a 1H NMR singlet at δ 9.85.
1. Assign the structure of A from the HRMS, IR, and 1H NMR data. Show how the coupling constants and molecular formula constrain the structure, and explain the small extra doublet.
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HRMS gives C11H12O2 and DBE = (2×11 + 2 − 12)/2 = 6. Four DBE are the para-disubstituted benzene ring, one is the carbonyl, and one is the alkene. IR 1665 cm−1 is a conjugated ketone; 1600 and 1510 cm−1 are aromatic. The two 2H doublets (J = 8.8 Hz) require a 1,4-disubstituted ring, the 3H singlets are ArOCH3 and COCH3, and the 1H doublets at 7.48 and 6.60 with J = 16.2 Hz are a trans alkene. A is (E)-4-(4-methoxyphenyl)but-3-en-2-one. The minor 12.4 Hz doublet is the Z isomer; its smaller coupling is characteristic of cis, not a second compound with a different formula.
2. Identify B and explain its retention order relative to A on silica.
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B has [M]+ 136.0524, corresponding to C8H8O2 (calc. 136.0524), and δ 9.85 is an aldehyde proton. It is unreacted 4-methoxybenzaldehyde. It elutes later (lower Rf) than the enone on silica because the aldehyde is more strongly retained by hydrogen bonding/dipole interactions with silanol groups; the enone is more hydrocarbon-like and less retained. This also explains why the two are separable in hexane/EtOAc 4:1 but not in a very polar solvent.
3. If the 12.4 Hz doublet integrates as 0.08 relative to the 16.2 Hz trans signal, estimate the isomeric purity and propose a reasonable purification check.
Read the worked answer
If the 12.4 Hz doublet integrates as 0.08 relative to the 16.2 Hz trans signal (1.00), then Z fraction = 0.08/(1.00 + 0.08) = 0.074, so the sample is about 92.6% E and 7.4% Z. Recrystallization can enrich the major isomer only if the Z isomer is more soluble in the chosen solvent; the result must be checked by NMR, not assumed. Repeated recrystallization from hot ethanol, or careful rechromatography, is reasonable. Do not report the product as pure E until the 12.4 Hz signal is below the detection limit; if it persists, consider a different recrystallization solvent or isomerization/equilibration conditions.